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A loss function $L(\theta ,a)$ measures the cost of taking action $a$ at parameter $\theta$. A decision rule $\delta$ selects an action from the data; its risk function is $R(\theta ,\delta )=E_{\theta }L(\theta ,\delta (X))$. A Bayes estimator minimizes the prior-averaged risk, or equivalently minimizes posterior expected loss at almost every observation. An admissible rule has no competing rule of no greater risk everywhere and strictly smaller risk somewhere. To construct a Bayes rule, form the posterior and minimize its expected loss over the allowed actions. Under squared error and finite posterior second moments the unique action is the posterior mean.

Put $S=|X{|}^2$. The given fourth moment and independence give $E_{v}S=nv$, $E_v{S}^2=n(n+2)v^2$. ThusEvery $\alpha \ne 1/(n+2)$ is strictly dominated at every $v>0$ by that coefficient. Hence it cannot be a finite-risk Bayes rule for any proper prior.

To make the second requested comparison, let $m_j=E_{\pi }{v}^j$ and assume finite quadratic Bayes risk. For the affine rules the integrated risk isWith $\alpha$ fixed, the minimum occurs at $\beta =(1-n\alpha )m_1$. Since $m_1>0$, this strictly improves the homogeneous rule unless $\alpha =1/n$, reducing risk by $(1-n\alpha {)}^2{m}_1^2$. Therefore a homogeneous rule with $\alpha \ne 1/n$ is not Bayes, as requested. Combining the two restrictions excludes every homogeneous coefficient.

For a proper prior without global finite moments, the same nonvacuous conclusion follows from the posterior-mean condition. LetFor $s>0$, $M_1^{\prime }(s)=-m(s)/2$. If the posterior mean were $\alpha s$, positivity requires $\alpha >0$ and $M_1=\alpha sm$. Differentiation forces $m(s)=C{s}^{-1-1/(2\alpha )}$. But the marginal density of $S$ would then be a constant times $s^{n/2-2-1/(2\alpha )}$, a pure power not integrable at both zero and infinity. This contradicts a proper prior's normalized predictive distribution. This is the proper-prior obstruction to homogeneous variance Bayes rules.

The finite-loss qualification matters: allowing all posterior losses to equal infinity makes every action a vacuous minimizer. For example, with $n=1$ and proper density $f_0(v)=(1+v{)}^{-2}$, the posterior second moment diverges at positive observations, so a literal infinite-loss definition would defeat the printed assertion. Ordinary meaningful Bayes rules exclude that convention.

Let $m=\int tdF(t)$ be the finite mean service time. Poisson arrivals see the stationary occupancy distribution, so the accepted rate is $\lambda (1-\pi )$. An accepted arrival's independent service time still has mean $m$. Little's law applied to occupied servers gives $\boxed{L=\lambda (1-\pi )m}$; it does not require exponential service times.

For the cafe, $\lambda =3/20$ and the mean table-occupation time is $(2/3)20+(1/3)30=70/3$. Thus $L=(7/2)(1-\pi )$ and the long-run revenue rate, at $2/5$ per occupied table per unit time, isCouple the cafe to an infinite-table system using the same arrivals and assigned service durations. Each accepted finite-cafe party is also present in the infinite system for exactly the same interval, so finite occupancy never exceeds infinite occupancy. Singles and pairs form independent thinned Poisson arrival streams of rates $1/10$ and $1/20$. Their infinite-server occupancies are independent Poisson variables of means two and $3/2$, hence their total is $N\sim \mathrm{Poisson}(7/2)$. Consequently $\boxed{\pi \le P(N\ge 23)}$.

The exponential Markov bound $P(N\ge r)\le \exp [\mu (e^t-1)-tr]$ is minimized at $t=\log (r/\mu )$. With $\mu =7/2,r=23$ it is $e^{-7/2}(7e/46{)}^{23}<5\times 10^{-11}$. This is far below $10^{-3}$, so blocking and the associated revenue loss are negligible in the specified model.

A meromorphic differential on a Riemann surface has local expression $\omega =a(z)dz$ with a meromorphic coefficient. On overlapping coordinates it transforms by $a(z)dz=\tilde{a}(w)dw$. For a meromorphic function $f$, the coefficients $f^{\prime }(z)$ obey this transformation by the chain rule, so $df$ is such a differential. Two nonzero differentials have coefficient ratio $a/b$; the derivative factors cancel on overlaps, yielding a global meromorphic function $h$ with $\eta =h\omega$.

An invertible holomorphic coordinate change has a nonzero holomorphic derivative, so multiplying by its derivative changes no local zero or pole order. Thus the differential's divisor is well defined. For a compact connected surface, the canonical-degree theorem states ${\sum }_P{\mathrm{ord}}_P\omega =2g-2$, including negative orders for poles. Therefore $\boxed{g=1+\frac{1}{2}\sum _P{\mathrm{ord}}_P\omega }$.

The projective closure has equation $U^2-V^2-W^2=0$. Its gradient $(2U,-2V,-2W)$ cannot vanish at a projective point, so it is nonsingular. At infinity $W=0$ gives the two points $[1:1:0]$ and $[1:-1:0]$. In the supplied sphere parameter,This has double poles at $t=\pm 1$, the two infinity points, and simple zeros at $t=\pm i$. At the parameter infinity use $s=1/t$, giving $v=2s/(1-s^2)$ and $dv=2(1+s^2)(1-s^2{)}^{-2}ds$, which is regular and nonzero at $s=0$. Hence $dv$ extends meromorphically over the whole projective curve. Its order sum $2-4=-2$ gives genus zero, consistently with the stated biholomorphism from the sphere.

The continuous dual space $X^{\ast }$ consists of bounded complex-linear functionals $\varphi :X\to \mathbb{C}$, with $\Vert \varphi \Vert ={\sup }_{\Vert x\Vert \le 1}|\varphi (x)|$. For $1\le s<\infty$, ${\ell }^s$ consists of sequences with ${\sum }_j|x_j{|}^s<\infty$, norm $({\sum }_j|x_j{|}^s{)}^{1/s}$; ${\ell }^{\infty }$ consists of bounded sequences with supremum norm.

For $a\in {\ell }^p$, define ${\varphi }_a(x)={\sum }_j{a}_j{x}_j$ on ${\ell }^q$. Hölder's inequality gives absolute convergence and $\Vert {\varphi }_a\Vert \le \Vert a{\Vert }_p$. For a finite truncation, choose $x_j=\overline{a_j}|a_j{|}^{p-2}/({\sum }_{j\le N}|a_j{|}^p{)}^{1/q}$ for $j\le N$, zero elsewhere, interpreting the numerator as zero when $a_j=0$. Its $q$-norm is one and its pairing equals $({\sum }_{j\le N}|a_j{|}^p{)}^{1/p}$. Letting $N$ grow proves equality of the norms.

Conversely, for a bounded $\varphi$ let $a_j=\varphi (e_j)$. The same finite test vector gives ${\sum }_{j\le N}|a_j{|}^p\le \Vert \varphi {\Vert }^p$, so $a\in {\ell }^p$. Finite truncations converge in ${\ell }^q$, and continuity gives $\varphi (x)={\sum }_j{a}_j{x}_j$. Thus $\boxed{({\ell }^q{)}^{\ast }\cong {\ell }^p\text{ isometrically}.}$

For ${\ell }^1$, boundedness gives $|a_j|\le \Vert \varphi \Vert$, while any bounded sequence defines an absolutely convergent pairing bounded by $\Vert a{\Vert }_{\infty }\Vert x{\Vert }_1$. Density of finite sequences again proves surjectivity, and single-coordinate tests give norm equality. Hence $\boxed{({\ell }^1{)}^{\ast }\cong {\ell }^{\infty }\text{ isometrically}.}$ The reversed argument fails for ${\ell }^{\infty }$: finite sequences are not dense in its supremum norm; their closure is $c_0$. For example the limit functional on convergent sequences extends by the Hahn-Banach theorem to ${\ell }^{\infty }$, vanishes on every $e_j$, yet takes value one on the constant-one sequence. It cannot be represented by pairing with an ${\ell }^1$ coefficient sequence.

The absence of $n$-simplices implies absence of every higher-dimensional simplex, since any such simplex would have an $n$-face. Therefore $X$ has a CW structure of dimension at most $n-1$. Give $S^n$ its CW structure with one zero-cell and one $n$-cell. The cellular approximation theorem states that any map of CW complexes is homotopic to a cellular map, which sends each source skeleton into the corresponding target skeleton. For $n\ge 1$, all target skeletons below dimension $n$ are the single zero-cell. The approximating map therefore sends all of $X$ there. Hence every given map is null-homotopic. This argument works for an infinite simplicial complex too; it does not require a finite subdivision or a finite collection of image simplices.

Let $p_1,\dots ,p_a$ be the first $a$ primes, and let $\varphi (x,a)$ count positive integers at most $x$ not divisible by any of them. The inclusion-exclusion principle givesFor $a=\pi (\sqrt{x})$, every composite at most $x$ is removed, so the survivors are one and the primes larger than $\sqrt{x}$. Thus the Legendre prime-counting formula is $\boxed{\pi (x)=\varphi (x,\pi (\sqrt{x}))+\pi (\sqrt{x})-1}$. This is the prime-counting formula, distinct from the factorial-valuation formula of the same name.

Fix $y\ge 2$ and put $a=\pi (y)$. Every prime exceeding $y$ is among the unsieved integers. Replacing each floor in the finite sum by its argument makes an error bounded by $2^a$, givingTake $x\to \infty$ with $y$ fixed. The supplied product estimate gives $0\le \mathrm{limsup}\pi (x)/x\le 1/\log y$. Since $y$ can be arbitrarily large, $\boxed{\underset{x\to \infty }{\lim }\pi (x)/x=0}$. The order of these limits avoids treating the growing inclusion-exclusion error as negligible.

Modulo two the quartic has no root, and the only irreducible monic quadratic is $x^2+x+1$. Reducing the quartic modulo that quadratic gives remainder one. Hence the quartic is irreducible over ${\mathbb{F}}_2$. Its splitting field is ${\mathbb{F}}_{16}$, and $\boxed{{\mathrm{Gal}}_{{\mathbb{F}}_2}\cong C_4}$, generated by Frobenius.

Modulo three it factors as $(x-1)(x^3+x^2+x+2)$. The cubic has no root at zero, one or two and is irreducible, so its splitting field is ${\mathbb{F}}_{27}$ and $\boxed{{\mathrm{Gal}}_{{\mathbb{F}}_3}\cong C_3}$. Both reductions are square-free. Irreducibility modulo two makes the rational quartic irreducible, and its rational Galois group is transitive on four roots. The unramified factorization theorem supplies a four-cycle from the modulo-two reduction and a three-cycle from the modulo-three reduction. Its order is divisible by twelve and divides twenty-four. An order-twelve subgroup would have index two and equal $A_4$, which has no four-cycle. Thus $\boxed{{\mathrm{Gal}}_{\mathbb{Q}}\cong S_4}$.

For fixed $k$-element subsets $U,V$, their rectangle is empty with probability $(1-p{)}^{k^2}$. A union bound over at most $n^{2k}$ choices gives $P(A)\le n^{2k}(1-p{)}^{k^2}$. For a specified vertex and a specified $(n-k)$-subset, the chance of no edges is $(1-p{)}^{n-k}$. There are at most $n\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le n^{k+1}$ choices on each side, givingFor $n\ge 6$, $k=\lceil \sqrt{n}\rceil \le n/2$ and $k^2\ge n$. Hence $P(A\cup B\cup C)<3n^{2k}(1-p{)}^{n/2}$. The logarithm of this bound is $\log 3+2k\log n+(n/2)\log (1-p)$, which tends to $-\infty$ because $\sqrt{n}\log n=o(n)$. Thus the bad-event probability tends to zero.

The printed phrase “almost surely” must mean asymptotically almost surely as $n\to \infty$. At any fixed $n$, the empty graph has positive probability $(1-p{)}^{n^2}$, so the literal probability-one assertion is false.

When no bad event occurs, verify Hall's marriage theorem. If $0<|U|\le k$, each vertex has more than $k$ neighbors because event $B$ is absent, so $|N(U)|\ge |U|$. If $k<|U|\le n-k$, a failure would leave at least $k$ vertices outside $N(U)$, producing a forbidden empty $k$-square. If $|U|>n-k$, every vertex in $Y$ meets $U$ because event $C$ is absent; hence $N(U)=Y$. The empty set also satisfies Hall. Therefore the graph has a perfect matching whenever the three events are absent, proving the random bipartite matching from empty-rectangle exclusion result with probability tending to one.

The Axiom of foundation states that every nonempty set $A$ has a member $a$ disjoint from $A$. The principle of membership induction states that a property $P$ satisfying $\forall x[(\forall y\in xP(y))\Rightarrow P(x)]$ holds for every set.

Assume Foundation and suppose a progressive property fails at a set $x$. In the transitive closure of $\{x\}$, form the nonempty set of counterexamples by Separation. Foundation selects a counterexample $a$ with no counterexample among its members. Transitivity makes those members belong to the closure, so each satisfies $P$, and progressiveness gives $P(a)$, a contradiction. Conversely, if a nonempty $A$ had no member disjoint from it, the property $P(x):x\notin A$ would be progressive: a member of $A$ necessarily has a member in $A$. Membership induction would then say no set belongs to $A$, a contradiction. Thus Foundation and membership induction are equivalent.

Apply induction to the property that a set belongs to some stage of the cumulative hierarchy. For each member $y$ of $x$, choose its least stage ordinal. Replacement collects these ordinals into a set; take an ordinal $\gamma$ large enough that all members lie in $V_{\gamma }$. Then $x\subseteq V_{\gamma }$ and $x\in V_{\gamma +1}$. This proves every set appears in the hierarchy.

For finite stages let $a_n=|V_n|$. Then $\boxed{a_0=0,a_{n+1}=2^{a_n}}$, giving $0,1,2,4,16,65536,\dots$. The stage $V_{\omega }$ is countably infinite, as the union of the finite stages. Thus $|V_{\omega +1}|=2^{{\aleph }_0}=|\mathbb{R}|$. Cantor's theorem makes $|V_{\omega +2}|$ strictly larger, and every later stage contains it. Earlier stages have at most countable size. Therefore $\boxed{|V_{\alpha }|=|\mathbb{R}|⟺\alpha =\omega +1}$.